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Clarification: Why the gauge symmetry of pure Yang-Mills is $PU(n)$ and not $SU(n)$?

+ 2 like - 0 dislike
217 views

I am quoting the following from the Wikipedia article on the projective unitary group:

In the pure Yang–Mills $SU(n)$ gauge theory, which is a gauge theory with only gluons and no fundamental matter, all fields transform in the adjoint of the gauge group $SU(n)$. The $Z/n$ center of $SU(n)$ commutes, being in the center, with $SU(n)$-valued fields and so the adjoint action of the center is trivial. Therefore the gauge symmetry is the quotient of $SU(n)$ by $Z/n$, which is $PU(n)$ and it acts on fields using the adjoint action described above.

In this context, the distinction between $SU(n)$ and $PU(n)$ has an important physical consequence. $SU(n)$ is simply connected, but the fundamental group of $PU(n)$ is $Z/n$, the cyclic group of order $n$. Therefore a $PU(n)$ gauge theory with adjoint scalars will have nontrivial codimension 2 vortices in which the expectation values of the scalars wind around $PU(n)$'s nontrivial cycle as one encircles the vortex. These vortices, therefore, also have charges in $Z/n$, which implies that they attract each other and when $n$ come into contact they annihilate. An example of such a vortex is the Douglas–Shenker string in $SU(n)$ Seiberg–Witten gauge theories.

  1. What is the center of $SU(n)$?
  2. What does it mean the adjoint action to be trivial? Which action are they talking about?
  3. I struggle to understand why the resulting gauge symmetry is $PU(n)$.
  4. What are the "codimension 2 vertices" of the scalars?
  5. Does this apply for pure $SU(3)$ QCD?
  6. References?
This post imported from StackExchange Physics at 2015-03-23 13:03 (UTC), posted by SE-user Marion
asked Mar 21, 2015 in Theoretical Physics by Marion Edualdo (250 points) [ no revision ]
You would probably do well to split up this question into multiple smaller ones. It seems awfully broad as of right now.

This post imported from StackExchange Physics at 2015-03-23 13:03 (UTC), posted by SE-user Danu
Why is it broad? I specifically mention where I am confused.

This post imported from StackExchange Physics at 2015-03-23 13:03 (UTC), posted by SE-user Marion
The fact that there are 6 subquestions seemed indicative to me, but maybe I'm wrong. Someone with a better understanding than me may be able to properly answer it! Please don't take the vote-to-close personally, by the way :)

This post imported from StackExchange Physics at 2015-03-23 13:03 (UTC), posted by SE-user Danu
Did you vote to close? You prefer it closed than answered?

This post imported from StackExchange Physics at 2015-03-23 13:03 (UTC), posted by SE-user Marion
Like I explained above, I think that the question, in its current form, is a bit too broad. This does not mean that I don't want to see it answered.

This post imported from StackExchange Physics at 2015-03-23 13:03 (UTC), posted by SE-user Danu
1. and 2. are trivial math questions, 3. is answered in your quote, 4. is actually a physics question, and for 5. you have to ask yourself the question if the center acts trivially on all objects in QCD. I agree that this is too broad, and would advise you to actually only ask 4.

This post imported from StackExchange Physics at 2015-03-23 13:03 (UTC), posted by SE-user ACuriousMind
I am not sure why $Z_n$ is the center of $SU(n)$ since, for example a $U(1)$ within $SU(n)$ should commute with all elements of $SU(n)$. Additionally, I am not sure how a $Z_n$ transformation acts on QCD fields. A $SU(3)$ transformation is just some matrix acting on quarks or gluons. But $Z_n$?

This post imported from StackExchange Physics at 2015-03-23 13:03 (UTC), posted by SE-user Marion

1 Answer

+ 2 like - 0 dislike

$SU(N)$ is the $N$-fold cover of $PSU(N)$. They share the same Lie algebra, so the Yang-Mills action would look identical locally. The center of $SU(N)$ is just $Z_N$. At the level of representations, the fundamental representation of $SU(N)$ is a projective representation of $PU(N)$, and only the adjoint ones are linear representations of $PU(N)$.

If the matter fields all transform in the adjoint representation, then it makes sense to say that the gauge group is actually $PU(N)$. A simple explanation is that by taking tensor product of adjoint representations you never get the fundamental ones, so the Hilbert space is restricted.

Because $PSU(N)=SU(N)/Z_N$, the global topology of $PU(N)$ is nontrivial. For example, the fundamental group $\pi_1(PU(N))=Z_N$, so there are nontrivial "vortex lines" in the scalar matter field, around which you pick up a holonomy in the center $Z_N$. These topological excitations themselves are one-dimensional objects, and have "codimension" 2.

Quarks in $SU(3)$ QCD transform as the fundamental representation.

This post imported from StackExchange Physics at 2015-03-23 13:03 (UTC), posted by SE-user Meng Cheng
answered Mar 22, 2015 by Meng (550 points) [ no revision ]
Hi and thanks. Can you provide me with some reference with the context you provide above (i.e. with focus on gauge theories)?

This post imported from StackExchange Physics at 2015-03-23 13:03 (UTC), posted by SE-user Marion

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